What about the convergence of : $I(z)=\int_{[0,z]}{(e^{-t²})}^{\text{erf(t)}}dt$ and is it entire function ??

Question

The copy of my question for reals posted here, Really i'm interesting to know about the convergence of $I(z)=\int_{[0,z]}{(e^{-t²})}^{\text{erf(t)}}dt$ for $z$ as a complex number and what about it's analyticity ? by other way is it entire function since $\text{erf(z)}$ is entire function ?And is it true that is always positive has no zero for non-zero $z$ then :$I(|z|) >0$ for $z\neq 0$ ?

Edit I have edited the question adding other question which discussed zero of this function in complex plane .

Note:$[0,z]$ denote the straight-line path from $0$ to $z$

Answer 1

A less confusing way to write this would be $\displaystyle\int^z_0e^{-t^2\operatorname{erf}(t)}\,dt$, and it's rather obvious it's an entire function, because it has a complex derivative everywhere. Since the integrand is analytic, the path of integration isn't relevant.
$I(z)>0$ doesn't make sense for complex $z$, and there's no reason to think $I(z)\neq0$ for $z\neq0$: the error function definitely has infinitely many complex zeros.

Answer 2

As Professor Vector's answer explains, your $I(z)$ is an entire function. You then ask

And is it true that is always positive has no zero for non-zero $z$ then :$I(z)>0$ for $z \neq 0$ ?

Positivity: clearly false. "Positive" is only sensible for real valued functions. $I$ is not constant so cannot be a real valued function (explanation below). For instance, $I(\mathrm{i}) = -0.305\dots{} + \mathrm{i}\,0.854\dots$. So it is impossible for $I$ to be positive.

Non-zero: clearly false; $I$ is not constant so cannot be bounded away from zero (explanation below). In fact, $I$ has a zero near $-0.814\dots{} + \mathrm{i}\, 1.930\dots$.

Liouville's theorem: A bounded entire function is constant.

Corollary: A non-constant entire function cannot be (only) real valued. Let $f$ be entire. Then $\mathrm{e}^{\mathrm{i}f}$ is entire. If $f$ is real valued, $\mathrm{e}^{\mathrm{i}f}$ is bounded (by $1$), so is constant, forcing $f$ constant.

Corollary: A non-constant entire function cannot be bounded away from zero. Let $f$ be entire and bounded away from $0$, $|f| > \varepsilon > 0$. Then $1/f$ is entire and bounded by $1/\varepsilon$, so is constant, forcing $f$ constant.